Here the sequence is defined by $a_1=1,a_2=2$, and $a_{k+1}=a_{k}+a_{k-1},\,\forall k\geq2$.
Stuck on this proof, this is what I have so far
Proof:
Basis step $(n=3): 2+1 \geq (\frac{3}{2})^{1}$
Assume for any arbitrary integer $k \geq 3, a_k \geq (\frac{3}{2})^{k-2}$
Need to show that $a_{k+1} \geq (\frac{3}{2})^{k-1}$
$a_{k+1} = a_k + a_{k-1}$
$ (\frac{3}{2})^{k-2} + a_{k-1} \geq (\frac{3}{2})^{k-1}$
$ (\frac{3}{2})^{k-2} + (\frac{3}{2})^{k-3}$
Base case: n = 2 holds.
Assume $a_{k} > \left (\frac{3}{2} \right) ^{k-2} $ for all $k \leq n$.
Then $a_{n + 1} = a_{n} + a_{n -1} > \left (\frac{3}{2} \right) ^{n-2} + \left (\frac{3}{2} \right) ^{n-3} = \left (\frac{3}{2} \right) ^{n-2} \left( \frac{3 + 2}{3} \right) = \left (\frac{3}{2} \right) ^{n-2} \cdot \frac{10}{6} > \left (\frac{3}{2} \right) ^{n-2} \cdot \frac{9}{6} = \left (\frac{3}{2} \right) ^{(n +1)-2}$.